AP Calculus AB Exam Question:

A ladder is sliding down a vertical wall. At the instant when the top of the ladder is 10 feet above the ground and sliding at a rate of 2 feet per second, the bottom of the ladder is 8 feet away from the wall and moving towards it at a rate of 3 feet per second.

At this instant, find the rate at which the angle between the ladder and the ground is changing. Round your answer to the nearest hundredth.

Step-by-Step Solution:

Let's label the pertinent variables:

• Let $x$ represent the distance between the bottom of the ladder and the wall.
• Let $y$ represent the height of the ladder on the wall.
• Let $\theta$ represent the angle between the ladder and the ground.

The problem provides:

• $\frac{{dx}}{{dt}} = -3$ ft/s (since the bottom of the ladder is moving towards the wall)
• $\frac{{dy}}{{dt}} = -2$ ft/s (since the top of the ladder is sliding down)
• $x = 8$ ft and $y = 10$ ft

We need to find $\frac{{d\theta}}{{dt}}$ when $x = 8$ ft and $y = 10$ ft.

Using trigonometric relationships, we know that:

$\cos(\theta) = \frac{{x}}{{\sqrt{{x^2 + y^2}}}}$ and $\sin(\theta) = \frac{{y}}{{\sqrt{{x^2 + y^2}}}}$

Differentiating both sides of these equations with respect to time $t$, we obtain:

$-\sin(\theta) \cdot \frac{{d\theta}}{{dt}} = \frac{{\frac{{dx}}{{dt}} \cdot \sqrt{{x^2 + y^2}} - \frac{{x}}{{\sqrt{{x^2 + y^2}}}} \cdot \frac{{d\sqrt{{x^2 + y^2}}}}{{dt}}}}{{x^2 + y^2}}$

Let's solve for $\frac{{d\theta}}{{dt}}$ by substituting the given values:

At $x = 8$ ft and $y = 10$ ft, we have:

$\frac{{dx}}{{dt}} = -3$ ft/s $\frac{{dy}}{{dt}} = -2$ ft/s $x = 8$ ft $y = 10$ ft

Substituting these values into the equation, we get:

$-\sin(\theta) \cdot \frac{{d\theta}}{{dt}} = \frac{{-3 \cdot \sqrt{{8^2 + 10^2}} - 8 \cdot \frac{{6x + 8y}}{{\sqrt{{8^2 + 10^2}}}}}}{{8^2 + 10^2}}$

$-\sin(\theta) \cdot \frac{{d\theta}}{{dt}} = \frac{{-3 \cdot 17.088 - 8 \cdot \frac{{6 \cdot 8 - 8 \cdot 10}}{{17.088}}}}{{164}}$

$-\sin(\theta) \cdot \frac{{d\theta}}{{dt}} = \frac{{-51.264 - 8 \cdot \frac{{-16}}{{17.088}}}}{{164}}$

$-\sin(\theta) \cdot \frac{{d\theta}}{{dt}} = \frac{{-51.264 + 8 \cdot \frac{{16}}{{17.088}}}}{{164}}$

$-\sin(\theta) \cdot \frac{{d\theta}}{{dt}} = \frac{{-51.264 + \frac{{128}}{{17.088}}}}{{164}}$

$\sin(\theta) \cdot \frac{{d\theta}}{{dt}} = \frac{{51.264 - \frac{{128}}{{17.088}}}}{{164}}$

$\frac{{d\theta}}{{dt}} = \frac{{\frac{{51.264 - \frac{{128}}{{17.088}}}}{{164}}}}{{\sin(\theta)}}$

Now, we can determine the value of $\sin(\theta)$ using the right triangle formed by the ladder, wall, and ground:

$\sin(\theta) = \frac{{y}}{{\sqrt{{x^2 + y^2}}}} = \frac{{10}}{{\sqrt{{8^2 + 10^2}}}}$

$\sin(\theta) = \frac{{10}}{{\sqrt{{164}}}}$

Now, substituting this value back into the equation, we have:

$\frac{{d\theta}}{{dt}} = \frac{{51.264 - \frac{{128}}{{17.088}}}}{{164}} \div \frac{{10}}{{\sqrt{{164}}}}$

$\frac{{d\theta}}{{dt}} = \frac{{51.264 - \frac{{128}}{{17.088}}}}{{164}} \cdot \frac{{\sqrt{{164}}}}{{10}}$

$\frac{{d\theta}}{{dt}} \approx 0.218$ rad/s (rounded to the nearest hundredth)

Therefore, the rate at which the angle between the ladder and the ground is changing is approximately $0.218$ rad/s.